Question

Evaluate:-
$\frac{tan^3\theta}{1+tan^2\theta}+\frac{cot^3\theta}{1+cot^2\theta}=\frac{1-2sin^2\theta.cos^2\theta}{sin\theta.cos\theta}$

Answer 1

Step-by-step explanation:

I am considering θ as A for the better understanding.

LHS:

$Given:\frac{tan^3A}{1+tan^2A}+\frac{cot^3A}{1+cot^2A}$

$=\frac{tan^3A}{sec^2A}+ \frac{cot^3A}{cosec^2A}$

$=\frac{\frac{sin^3A}{cos^3A}}{\frac{1}{cos^2A}}+\frac{\frac{cos^3A}{sin^3A}}{\frac{1}{sin^2A}}$

$=\frac{sin^3A}{cosA}+\frac{cos^3A}{sinA}$

$=\frac{sin^4A + cos^4A}{sinA*cosA}$

$=\frac{sin^4A+cos^4A+2sin^2Acos^2A-2sin^2Acos^2A}{sinAcosA}$

$=\frac{(sin^2A+cos^2A)^2-2sin^2Acos^2A}{sinAcosA}$

$=\frac{1-2sin^2Acos^2A}{sinAcosA}$

RHS:

Hope it helps!

Answer 2

Solution :

$\frac{tan^3\theta}{1+tan^2\theta}+\frac{cot^3\theta}{1+cot^2\theta}\\\\\implies \frac{tan^3\theta}{sec^2\theta}+\frac{cot^3\theta}{cosec^2\theta}\\\\\implies \frac{\frac{sin^3}{cos^3\theta}}{sec^2\theta}+\frac{\frac{cos^3}{sin^3\theta}}{cosec^2\theta}$

$\implies \frac{sin^3\theta}{cos^3\theta sec^2\theta}+\frac{cos^3\theta}{cosec^2\thetasin^3\theta}\\\\\implies \frac{sin^3\theta}{cos\theta}+\frac{cos^3\theta}{sin\theta}$

$\implies \frac{sin^4\theta+cos^4\theta}{sin\thetacos\theta}\\\\\implies \frac{(sin^2\theta+cos^2\theta)^2-2sin^2\theta cos^2\theta}{sin\theta cos\theta}$

$\implies \frac{1-2sin^2\theta cos^2\theta}{sin\theta cos\theta}$

NOTE :

$\bigstar tan^3\theta=\frac{sin^3\theta}{cos^3\theta}\\\\\bigstar cot^3\theta=\frac{cos^3\theta}{sin^3\theta}\\\\\bigstar 1+tan^2\theta=sec^2\theta\\\\\bigstar 1+cot^2\theta=cosec^2\theta$

$\bigstar sin^2\theta+cos^2\theta=1\\\\\bigstar a^4+b^4 = (a^2)^2+(b^2)^2+2a^2b^2-2a^2b^2=(a^2+b^2)-2a^2b^2$